A system executes a cyclic process during which there are four heat transfers of 1000, 90, 700 and 250 J and four work interactions of 100, 50, 200 and W 4J. Find the magnitude and direction of the fourth work interaction, W 4.
1690 J
Understanding the First Law of Thermodynamics for a Cyclic Process
The First Law of Thermodynamics is a statement of the conservation of energy. For a system undergoing any process, the change in internal energy (\(\Delta U\)) is equal to the net heat added to the system (\(Q\)) minus the net work done by the system (\(W\)). Mathematically, this is expressed as:
\[ \Delta U = Q - W \]For a cyclic process, the system returns to its initial state. Since internal energy is a state property, the change in internal energy over a complete cycle is zero:
\[ \Delta U_{cycle} = 0 \]Applying the First Law to a cyclic process, we get:
\[ 0 = Q_{net} - W_{net} \] \[ Q_{net} = W_{net} \]This means that for a cyclic process, the net heat transfer into the system is equal to the net work transfer out of the system.
Applying the First Law to the Cyclic Process
The problem provides a system executing a cyclic process with four heat transfers and four work interactions. Let the heat transfers be \(Q_1, Q_2, Q_3, Q_4\) and the work interactions be \(W_1, W_2, W_3, W_4\). The given magnitudes are:
- Heat transfers: 1000 J, 90 J, 700 J, 250 J
- Work interactions: 100 J, 50 J, 200 J, \(W_4\) J
The net heat transfer (\(Q_{net}\)) is the sum of all heat transfers, and the net work transfer (\(W_{net}\)) is the sum of all work interactions. For the cyclic process, we have \(Q_{net} = W_{net}\).
Calculating Net Heat Transfer
The problem lists four heat transfers. To find the net heat transfer, we need to consider their signs. Assuming all listed heat transfers are positive (heat added to the system), the net heat transfer is:
\[ Q_{net} = Q_1 + Q_2 + Q_3 + Q_4 \] \[ Q_{net} = 1000 \, \text{J} + 90 \, \text{J} + 700 \, \text{J} + 250 \, \text{J} \] \[ Q_{net} = 2040 \, \text{J} \]Calculating Net Work Transfer
Similarly, we sum the work interactions. Assuming the first three listed work interactions are positive (work done by the system), the net work transfer is:
\[ W_{net} = W_1 + W_2 + W_3 + W_4 \] \[ W_{net} = 100 \, \text{J} + 50 \, \text{J} + 200 \, \text{J} + W_4 \] \[ W_{net} = 350 \, \text{J} + W_4 \]Finding the Fourth Work Interaction, \(W_4\)
Using the First Law for a cyclic process, \(Q_{net} = W_{net}\), we can set up the equation:
\[ 2040 \, \text{J} = 350 \, \text{J} + W_4 \]Now, we solve for \(W_4\):
\[ W_4 = 2040 \, \text{J} - 350 \, \text{J} \] \[ W_4 = 1690 \, \text{J} \]The magnitude of the fourth work interaction is 1690 J. The positive sign indicates that this work is done by the system (work output).
Summary of Calculations
| Interaction Type | Magnitude (J) | Assumed Sign | Value (J) |
|---|---|---|---|
| Heat 1 | 1000 | + (Added) | +1000 |
| Heat 2 | 90 | + (Added) | +90 |
| Heat 3 | 700 | + (Added) | +700 |
| Heat 4 | 250 | + (Added) | +250 |
| Net Heat (\(Q_{net}\)) | +2040 | ||
| Work 1 | 100 | + (By System) | +100 |
| Work 2 | 50 | + (By System) | +50 |
| Work 3 | 200 | + (By System) | +200 |
| Work 4 | \(W_4\) | ? | \(W_4\) |
| Net Work (\(W_{net}\)) | \(350 + W_4\) | ||
Equating Net Heat and Net Work for the cyclic process:
\[ Q_{net} = W_{net} \] \[ 2040 = 350 + W_4 \] \[ W_4 = 1690 \, \text{J} \]The fourth work interaction is 1690 J. The positive sign implies the direction is work done by the system.
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